Image of a character remains the same when restricting to a totally ramified extension
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Problem I want to prove: Let $chi: G_K to mathbb{C}^*$ be an unramified character and let $L/K$ be a cyclic totally ramified extension. Then $chi(G_K)=chi(G_L)$.
All I managed to do was considering all definitions and characterizations (without further success):
Definition 1: Let $G_K$ be the absolute Galois group of a local field $K$. We will call a group homomorphism $chi: G_K to mathbb{C}^*$ with finite image a character on $K$.
Remark: Since every finite subgroup of $mathbb{C}^*$ is cyclic, it is generated by a primitive root of unity. So in our case, every character $chi$ corresponds to a unique cyclic Galois extension $F/K$ of degree $n$, the cardinality of the image of $chi$, and an isomorphism
$$bar{chi}: operatorname{Gal}(F/K) xrightarrow{sim} langle xi_n rangle subseteq mathbb{C}^*$$
where $xi_n$ is a primitive $n$-th root of unity. We also say that $chi$ cuts out the extension $F/K$.
Definition 2: We call a character $chi: G_K to mathbb{C}^*$
unramified if the restriction of $chi$ to $F$ is the trivial map, i.e. $chi|_F(sigma)=1$ for all $sigma in G_F$, and
totally ramified if $chi(I_K) = chi(G_K)$ where $I_K$ denotes the inertia subgroup of $K$.
Remark A character $chi$ which cuts out an extension $F/K$ is unramified (resp. totally ramified) if and only if $F/K$ is unramified (resp. totally ramified). Another characterization for $chi$ being unramified (resp. totally ramified) is that the order of $chi(I_K)$ (also called the ramification index of $chi$) is equal to $1$ (resp. $[F:K]$) where $I_K$ denotes the inertia subgroup of $K$.
The intuitive approach for the Problem should somehow deal with the fact that the residue fields (resp. the inertia subgroups) remain the same when going from $K$ to $L$. But I am not able to proceed with the technical proof.
Could you please help me with my problem? Thank you in advance!
abstract-algebra group-theory algebraic-number-theory local-field ramification
asked Nov 20 at 0:38
Diglett
822420
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Problem I want to prove: Let $chi: G_K to mathbb{C}^*$ be an unramified character and let $L/K$ be a cyclic totally ramified extension. Then $chi(G_K)=chi(G_L)$.
All I managed to do was considering all definitions and characterizations (without further success):
Definition 1: Let $G_K$ be the absolute Galois group of a local field $K$. We will call a group homomorphism $chi: G_K to mathbb{C}^*$ with finite image a character on $K$.
Remark: Since every finite subgroup of $mathbb{C}^*$ is cyclic, it is generated by a primitive root of unity. So in our case, every character $chi$ corresponds to a unique cyclic Galois extension $F/K$ of degree $n$, the cardinality of the image of $chi$, and an isomorphism
$$bar{chi}: operatorname{Gal}(F/K) xrightarrow{sim} langle xi_n rangle subseteq mathbb{C}^*$$
where $xi_n$ is a primitive $n$-th root of unity. We also say that $chi$ cuts out the extension $F/K$.
Definition 2: We call a character $chi: G_K to mathbb{C}^*$
unramified if the restriction of $chi$ to $F$ is the trivial map, i.e. $chi|_F(sigma)=1$ for all $sigma in G_F$, and
totally ramified if $chi(I_K) = chi(G_K)$ where $I_K$ denotes the inertia subgroup of $K$.
Remark A character $chi$ which cuts out an extension $F/K$ is unramified (resp. totally ramified) if and only if $F/K$ is unramified (resp. totally ramified). Another characterization for $chi$ being unramified (resp. totally ramified) is that the order of $chi(I_K)$ (also called the ramification index of $chi$) is equal to $1$ (resp. $[F:K]$) where $I_K$ denotes the inertia subgroup of $K$.
The intuitive approach for the Problem should somehow deal with the fact that the residue fields (resp. the inertia subgroups) remain the same when going from $K$ to $L$. But I am not able to proceed with the technical proof.
Could you please help me with my problem? Thank you in advance!
abstract-algebra group-theory algebraic-number-theory local-field ramification
asked Nov 20 at 0:38
Diglett
822420
A definition of unramified character $chi$ is that $chi$ kills the inertia subgroup. Using this definition I think it should be obvious? $L/K$ being cyclic should be irrelevant. I'm also wondering if $F/K$ is necessarily cyclic? I feel you wanted to say that $F = K(zeta_n)$, which is not necessarily cyclic, isn't it?
– dyf
2 days ago
@dyf: Thank you for your response? "Using this definition (of yours) I think it should be obvious?" It might be but I do not see it immediately. Could you please elaborate this point?
– Diglett
2 days ago
@dyf: I also think that $L/K$ being cyclic is irrelevant, but I just mentioned it because it appears to be cyclic in one of my problems. "I'm also wondering if $F/K$ is necessarily cyclic?" I think this must be true because every finite subgroup of $mathbb{C}^*$ must be cyclic, otherwise we would not have the correspondence for the characters.
– Diglett
2 days ago
For the former, I thought if if $F/L/K$, then $I_{F/L} subset I_{F/K}$, now now that I think of it, I am not even sure this is right (sorry to disappoint you). For the latter, I see what you mean now, I subtly used Kronecker-Weber to think of $F$ as $K(zeta_n)$, but indeed I should not have done that. (Thanks!)
– dyf
2 days ago
add a comment |
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0
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up vote
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Problem I want to prove: Let $chi: G_K to mathbb{C}^*$ be an unramified character and let $L/K$ be a cyclic totally ramified extension. Then $chi(G_K)=chi(G_L)$.
All I managed to do was considering all definitions and characterizations (without further success):
Definition 1: Let $G_K$ be the absolute Galois group of a local field $K$. We will call a group homomorphism $chi: G_K to mathbb{C}^*$ with finite image a character on $K$.
Remark: Since every finite subgroup of $mathbb{C}^*$ is cyclic, it is generated by a primitive root of unity. So in our case, every character $chi$ corresponds to a unique cyclic Galois extension $F/K$ of degree $n$, the cardinality of the image of $chi$, and an isomorphism
$$bar{chi}: operatorname{Gal}(F/K) xrightarrow{sim} langle xi_n rangle subseteq mathbb{C}^*$$
where $xi_n$ is a primitive $n$-th root of unity. We also say that $chi$ cuts out the extension $F/K$.
Definition 2: We call a character $chi: G_K to mathbb{C}^*$
unramified if the restriction of $chi$ to $F$ is the trivial map, i.e. $chi|_F(sigma)=1$ for all $sigma in G_F$, and
totally ramified if $chi(I_K) = chi(G_K)$ where $I_K$ denotes the inertia subgroup of $K$.
Remark A character $chi$ which cuts out an extension $F/K$ is unramified (resp. totally ramified) if and only if $F/K$ is unramified (resp. totally ramified). Another characterization for $chi$ being unramified (resp. totally ramified) is that the order of $chi(I_K)$ (also called the ramification index of $chi$) is equal to $1$ (resp. $[F:K]$) where $I_K$ denotes the inertia subgroup of $K$.
The intuitive approach for the Problem should somehow deal with the fact that the residue fields (resp. the inertia subgroups) remain the same when going from $K$ to $L$. But I am not able to proceed with the technical proof.
Could you please help me with my problem? Thank you in advance!
abstract-algebra group-theory algebraic-number-theory local-field ramification
asked Nov 20 at 0:38
Diglett
822420
Problem I want to prove: Let $chi: G_K to mathbb{C}^*$ be an unramified character and let $L/K$ be a cyclic totally ramified extension. Then $chi(G_K)=chi(G_L)$.
All I managed to do was considering all definitions and characterizations (without further success):
Definition 1: Let $G_K$ be the absolute Galois group of a local field $K$. We will call a group homomorphism $chi: G_K to mathbb{C}^*$ with finite image a character on $K$.
Remark: Since every finite subgroup of $mathbb{C}^*$ is cyclic, it is generated by a primitive root of unity. So in our case, every character $chi$ corresponds to a unique cyclic Galois extension $F/K$ of degree $n$, the cardinality of the image of $chi$, and an isomorphism
$$bar{chi}: operatorname{Gal}(F/K) xrightarrow{sim} langle xi_n rangle subseteq mathbb{C}^*$$
where $xi_n$ is a primitive $n$-th root of unity. We also say that $chi$ cuts out the extension $F/K$.
Definition 2: We call a character $chi: G_K to mathbb{C}^*$
unramified if the restriction of $chi$ to $F$ is the trivial map, i.e. $chi|_F(sigma)=1$ for all $sigma in G_F$, and
totally ramified if $chi(I_K) = chi(G_K)$ where $I_K$ denotes the inertia subgroup of $K$.
Remark A character $chi$ which cuts out an extension $F/K$ is unramified (resp. totally ramified) if and only if $F/K$ is unramified (resp. totally ramified). Another characterization for $chi$ being unramified (resp. totally ramified) is that the order of $chi(I_K)$ (also called the ramification index of $chi$) is equal to $1$ (resp. $[F:K]$) where $I_K$ denotes the inertia subgroup of $K$.
The intuitive approach for the Problem should somehow deal with the fact that the residue fields (resp. the inertia subgroups) remain the same when going from $K$ to $L$. But I am not able to proceed with the technical proof.
Could you please help me with my problem? Thank you in advance!
abstract-algebra group-theory algebraic-number-theory local-field ramification
abstract-algebra group-theory algebraic-number-theory local-field ramification
asked Nov 20 at 0:38
Diglett
822420
asked Nov 20 at 0:38
Diglett
822420
edited 2 days ago
asked Nov 20 at 0:38
Diglett
822420
asked Nov 20 at 0:38
Diglett
822420
asked Nov 20 at 0:38
Diglett
822420
822420
A definition of unramified character $chi$ is that $chi$ kills the inertia subgroup. Using this definition I think it should be obvious? $L/K$ being cyclic should be irrelevant. I'm also wondering if $F/K$ is necessarily cyclic? I feel you wanted to say that $F = K(zeta_n)$, which is not necessarily cyclic, isn't it?
– dyf
2 days ago
@dyf: Thank you for your response? "Using this definition (of yours) I think it should be obvious?" It might be but I do not see it immediately. Could you please elaborate this point?
– Diglett
2 days ago
@dyf: I also think that $L/K$ being cyclic is irrelevant, but I just mentioned it because it appears to be cyclic in one of my problems. "I'm also wondering if $F/K$ is necessarily cyclic?" I think this must be true because every finite subgroup of $mathbb{C}^*$ must be cyclic, otherwise we would not have the correspondence for the characters.
– Diglett
2 days ago
For the former, I thought if if $F/L/K$, then $I_{F/L} subset I_{F/K}$, now now that I think of it, I am not even sure this is right (sorry to disappoint you). For the latter, I see what you mean now, I subtly used Kronecker-Weber to think of $F$ as $K(zeta_n)$, but indeed I should not have done that. (Thanks!)
– dyf
2 days ago
add a comment |
A definition of unramified character $chi$ is that $chi$ kills the inertia subgroup. Using this definition I think it should be obvious? $L/K$ being cyclic should be irrelevant. I'm also wondering if $F/K$ is necessarily cyclic? I feel you wanted to say that $F = K(zeta_n)$, which is not necessarily cyclic, isn't it?
– dyf
2 days ago
@dyf: Thank you for your response? "Using this definition (of yours) I think it should be obvious?" It might be but I do not see it immediately. Could you please elaborate this point?
– Diglett
2 days ago
@dyf: I also think that $L/K$ being cyclic is irrelevant, but I just mentioned it because it appears to be cyclic in one of my problems. "I'm also wondering if $F/K$ is necessarily cyclic?" I think this must be true because every finite subgroup of $mathbb{C}^*$ must be cyclic, otherwise we would not have the correspondence for the characters.
– Diglett
2 days ago
For the former, I thought if if $F/L/K$, then $I_{F/L} subset I_{F/K}$, now now that I think of it, I am not even sure this is right (sorry to disappoint you). For the latter, I see what you mean now, I subtly used Kronecker-Weber to think of $F$ as $K(zeta_n)$, but indeed I should not have done that. (Thanks!)
– dyf
2 days ago
A definition of unramified character $chi$ is that $chi$ kills the inertia subgroup. Using this definition I think it should be obvious? $L/K$ being cyclic should be irrelevant. I'm also wondering if $F/K$ is necessarily cyclic? I feel you wanted to say that $F = K(zeta_n)$, which is not necessarily cyclic, isn't it?
– dyf
2 days ago
A definition of unramified character $chi$ is that $chi$ kills the inertia subgroup. Using this definition I think it should be obvious? $L/K$ being cyclic should be irrelevant. I'm also wondering if $F/K$ is necessarily cyclic? I feel you wanted to say that $F = K(zeta_n)$, which is not necessarily cyclic, isn't it?
– dyf
2 days ago
@dyf: Thank you for your response? "Using this definition (of yours) I think it should be obvious?" It might be but I do not see it immediately. Could you please elaborate this point?
– Diglett
2 days ago
@dyf: Thank you for your response? "Using this definition (of yours) I think it should be obvious?" It might be but I do not see it immediately. Could you please elaborate this point?
– Diglett
2 days ago
@dyf: I also think that $L/K$ being cyclic is irrelevant, but I just mentioned it because it appears to be cyclic in one of my problems. "I'm also wondering if $F/K$ is necessarily cyclic?" I think this must be true because every finite subgroup of $mathbb{C}^*$ must be cyclic, otherwise we would not have the correspondence for the characters.
– Diglett
2 days ago
@dyf: I also think that $L/K$ being cyclic is irrelevant, but I just mentioned it because it appears to be cyclic in one of my problems. "I'm also wondering if $F/K$ is necessarily cyclic?" I think this must be true because every finite subgroup of $mathbb{C}^*$ must be cyclic, otherwise we would not have the correspondence for the characters.
– Diglett
2 days ago
For the former, I thought if if $F/L/K$, then $I_{F/L} subset I_{F/K}$, now now that I think of it, I am not even sure this is right (sorry to disappoint you). For the latter, I see what you mean now, I subtly used Kronecker-Weber to think of $F$ as $K(zeta_n)$, but indeed I should not have done that. (Thanks!)
– dyf
2 days ago
For the former, I thought if if $F/L/K$, then $I_{F/L} subset I_{F/K}$, now now that I think of it, I am not even sure this is right (sorry to disappoint you). For the latter, I see what you mean now, I subtly used Kronecker-Weber to think of $F$ as $K(zeta_n)$, but indeed I should not have done that. (Thanks!)
– dyf
2 days ago
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A definition of unramified character $chi$ is that $chi$ kills the inertia subgroup. Using this definition I think it should be obvious? $L/K$ being cyclic should be irrelevant. I'm also wondering if $F/K$ is necessarily cyclic? I feel you wanted to say that $F = K(zeta_n)$, which is not necessarily cyclic, isn't it?
– dyf
2 days ago
@dyf: Thank you for your response? "Using this definition (of yours) I think it should be obvious?" It might be but I do not see it immediately. Could you please elaborate this point?
– Diglett
2 days ago
@dyf: I also think that $L/K$ being cyclic is irrelevant, but I just mentioned it because it appears to be cyclic in one of my problems. "I'm also wondering if $F/K$ is necessarily cyclic?" I think this must be true because every finite subgroup of $mathbb{C}^*$ must be cyclic, otherwise we would not have the correspondence for the characters.
– Diglett
2 days ago
For the former, I thought if if $F/L/K$, then $I_{F/L} subset I_{F/K}$, now now that I think of it, I am not even sure this is right (sorry to disappoint you). For the latter, I see what you mean now, I subtly used Kronecker-Weber to think of $F$ as $K(zeta_n)$, but indeed I should not have done that. (Thanks!)
– dyf
2 days ago